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The exercise says like this

There are 3 boxes containing balls:

  • The first has 3 white and 2 black
  • The second has 2 white and 3 black
  • The third has 3 white and 3 black

Only one ball is chosen from each box. If two of the chosen balls are white:

  • What is the probability that one of the balls was chosen from the first box?

The result is 15/19

  • So, what's the real solution to this exercise?

The data I gather so far are

  • Ci: "The ball is of the box i" such that i = 1 or 2 or 3
  • B: "The ball is white"

  • N: "The ball is black"

    P(B|C1) = 3/5, P(N|C1) = 2/5

    P(B|C2) = 2/5, P(N|C2) = 3/5

    P(N|C3) = 1/2, P(B|C3) = 1/2

    P(Ci) = 1/3

    P(B|C1unionC2unionC3) = 2/3, Or should it be an intersection between the three?

    P(B) = P(B|C1)*P(C1) + P(B|C2)*P(C2) + P(B|C3)*P(C3)

    P(B) = 1/2

    P(C1|B) = P(B|C1)*P(C1) / P(B) = 2/5

  • 0
    Do you mean exactly two or at least two? – 2017-01-19
  • 0
    In _"(...) If two of the chosen balls are white:"_ I mean exactly 2. – 2017-01-19

2 Answers 2

3

There are three scenarios that yield exactly two white balls:

$$P(W,W,B)=\frac 35 \times \frac 25\times \frac 36= \frac 3{25}$$

$$P(W,B,W)=\frac 35 \times \frac 35\times \frac 36= \frac 9{50}$$

$$P(B,W,W)=\frac 25 \times \frac 25\times \frac 36= \frac 2{25}$$

The answer you seek is $$\frac {P(W,W,B)+P(W,B,W)}{P(W,W,B)+P(W,B,W)+P(B,W,W)}=\frac {6+9}{6+9+4}=\frac {15}{19}$$

  • 0
    Why in the last calculation do you divide the sum of the probabilities of favorable cases between the sum of the probabilities of the possible cases? What are you applying? Thanks. – 2017-01-20
  • 1
    Let $X$ be the event "the ball from the first urn is White". Let $Y$ be the event "exactly two of the three are white". The you want the conditional probability $P(X\,|\,Y)$ but this is $\frac {P(X\cap Y)}{P(Y)}$ and this is what I wrote. – 2017-01-20
  • 0
    But P(X∩Y) = P(Y|X)*P(X), or not? Why are you adding up P(W,W,B)+P(W,B,W)? – 2017-01-20
  • 0
    My sum is represents every case in which $X$ and $Y$ both happen! Nothing wrong with your expression, if you can compute it. – 2017-01-20
  • 0
    Excuse my ignorance. It is okay to think like... **X = [(W1 and W2 and B3) or (W1 and B2 and W3) or (B1 and W2 and W3)]** after **Y = [(W1 and W2 and B3) or (W1 and B2 and W3)]**, but **X and Y = Y**, then **P (Y / X) = P (Y) / P (X)**? Wi: "White box i", Bi: "Black box i" (I don't know why cannot write math symbols, sorry) – 2017-01-20
  • 0
    Not following. You appear to be including $(B,W,W)$ as an event in $X$ but it clearly isn't. In general, I'm not clear how you are looking at this. It's a finite problem! There are $8$ possible cases. Note that they are not equiprobable. Write all of them out and compute their probabilities. Which contribute to $X$? Which to $Y$? – 2017-01-20
  • 0
    Sorry, accidentally I have change the meaning of the events. In my comment **X** is _"Two of the three are white"_ and **Y** is _"One white of the two are from the box one"_ – 2017-01-20
  • 0
    But I can't understand anything in your comment. What on earth does "But $X$ and $Y=Y$" mean? Like I said, there are only $8$ possible outcomes here. Just do the problem by enumeration if you are confused. – 2017-01-20
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    I mean that X intersection Y equals Y. – 2017-01-20
  • 0
    But that is obviously not true. – 2017-01-20
  • 1
    Have you done the enumeration? List the $8$ cases and compute all the probabilities. Check that they add to $1$. Which events comprise (my) $X$? which comprise (my) $Y$? Which comprise $X\cap Y$? – 2017-01-20
  • 0
    X={(W,W,W);(W,W,B);(W,B,W);(W,B,B)}, Y={(W,W,B);(W,B,W);(B,W,W)}, X∩Y={(W,W,B);(W,B,W)}? – 2017-01-20
  • 0
    Yes. And now you just need to compute the probabilities of each case. – 2017-01-20
  • 0
    Ω = {(W,W,W);(W,W,B);(W,B,B);(B,B,B);(B,B,W);(B,W,W);(W,B,W);(B,W,B);}? – 2017-01-20
  • 0
    They are 3/25; 3/25; 9/50; 3/25; 3/25; 2/25; 9/50; 2/25 respectively. And the sum is 1. – 2017-01-20
  • 0
    Looks good. Note that those numbers match mine where I computed them. Now do you see how to compute $P(X\cap Y)$? – 2017-01-20
  • 0
    Now I can see it. Thanks very much. – 2017-01-20
1

This is an application of Baye's theorem.

Let $A$ be the event: exactly two white balls appear.

Let $B$ be the event: A white ball is extracted from the first bin.

We want to calculate $P(B|A)$.

By Baye's theorem it's equal to $\frac{P(A|B)P(B)}{P(A)}=\frac{(2/5\times 1/2+ 3/5\times 1/2)(3/5)}{P(A)}=\frac{3}{10P(A)}$.

So we need to calculate $P(A)$.

There are $3$ ways in which exactly $3$ balls can be extracted, we can add the probabilities to get: $3/5\times2/5\times 1/2+ 2/5\times 2/5\times 1/2+ 3/5\times 3/5\times 1/2=\frac{19}{50}$

So the final answer is $\frac{150}{190}=\frac{15}{19}$

  • 0
    Can you explain me why P(A/B) is equal to (2/5*1/2+3/5*1/2)? Thanks. – 2017-01-19
  • 0
    because there are two options, white,white,black and white,black, white. The first has probability 2/5*1/2 and the second 3/5*1/2 – 2017-01-19
  • 0
    But 2/5 is the probability of obtaining a white from the second, 3/5 is the probability of obtaining a white from the first, and 1/2 is the probability of obtaining a black (or white) from the third. Why you simply multiply the second by the third and the first by the third? – 2017-01-19
  • 0
    no no, because you are already assuming that you got white in the first one, and all $3$ draws are independent. In short: I am just multiplying second by third and adding, because we are working in the case in which the first choice is white. – 2017-01-19
  • 0
    In exercises like this: What are the advantages and disadvantages of simply applying the Laplace rule (instead of Bayes' theorem) after obtaining the probabilities of favorable and possible cases? Is there a difference? Thanks. – 2017-01-19
  • 0
    Laplace's rule of succession is a special case of Bayes theorem but you need all events to be identically distributed. It doesn't apply here because each selection has different probabilities. – 2017-01-19
  • 0
    So, why do we come up with the same result? Is there a coincidence? – 2017-01-19
  • 0
    wouldn't the Laplace rule give $\frac{2+1}{2+2}=\frac{3}{4}$? – 2017-01-19
  • 0
    Ok, so I don't know what Lulu applied. Thanks. – 2017-01-19